Question

The ratio of distances covered by a freely falling body (starting from rest) during the $1^{\text{st}},2^{\text{nd}},3^{\text{rd}}$ second of its motion is: [Take $g=10{\text{m/s}}^2$]

• A. $3:6:9$
• B. $1:3:5$
• C. $2:3:5$
• D. $1:3:7$

Answer 1

The correct option is B $1:3:5$

Considering vertically downwards as $+ve$ direction
$u=0,a=+g$
When acceleration is constant, displacement of a body in $n^{th}$ second is given by:
$s=u+\frac{a}{2}(2n-1)$
In free fall motion displacement $=$ distance covered
For $n=1$,
$s_1=0+\frac{1}{2}\times 10(2-1)=5\text{m}$
For $n=2$
$s_2=0+\frac{1}{2}\times 10\left(\right(2\times 2)-1)=15\text{m}$
Similarly for $n=3$
$s_3=0+\frac{1}{2}\times 10\left(\right(2\times 3)-1)=25\text{m}$
Therefore, ratio is
$s_1:s_2:s_3=5:15:25=1:3:5$